Question: A curve in the plane is defined parametrically by the equations $x=8e^{3t}$ and $y=\cos(4t)$. Find $\dfrac{dy}{dx}$. Choose 1 answer: Choose 1 answer: (Choice A) A $-\dfrac{3\cos(4t)+4\sin(3t)}{8e^{3t}}$ (Choice B) B $-\dfrac{\sin(4t)}{6e^{3t}}$ (Choice C) C $\dfrac{2e^{3t}}{\sin(4t)}$ (Choice D) D $-4\sin(4t)$
In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=8e^{3t}$ and $y=\cos(4t)$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\cos(4t)\right)}{\dfrac{d}{dt}(8e^{3t})} \\\\ &=\dfrac{-4\sin(4t)}{24e^{3t}} \\\\ &=-\dfrac{\sin(4t)}{6e^{3t}} \gray{\text{Simplify}} \end{aligned}$ In conclusion, $\dfrac{dy}{dx}=-\dfrac{\sin(4t)}{6e^{3t}}$.